00:01
With this question, we want to find the average occupation number of rosons.
00:06
Okay, so to do this question, this is how we, this is what we need to do.
00:20
Okay, so we are given that the tn is, okay, n times exponential of minus n e over kvt.
00:43
So n is the normalization factor.
00:57
So first thing we need to do is to find a normalization factor.
01:03
Before we find a mean occupation number.
01:08
So, and we know that summation of all the probability is equal to 1.
01:14
So this means that summation n times e minus n e over kvt equals to 1.
01:24
So this thing itself is a geometric series.
01:37
Alright, so we have a first term is one.
01:42
Okay, yeah, the summation is from zero to infinity.
01:49
Okay, so the first term is one, and then the common ratio is e to the minus e over kvt.
01:57
T so n times 1 over 1 minus e minus e over kb t is equal to 1.
02:09
So we can tell that the normalization factor is 1 minus e to the minus e over kb t and then let e to the minus e over kvt because to z so n equals to 1 minus z z.
02:29
Okay...